# A little geometry help, please
**URL:** https://boards.straightdope.com/t/a-little-geometry-help-please/96704
**Category:** Factual Questions
**Created:** [December 3, 2001, 10:06pm UTC](https://boards.straightdope.com/t/a-little-geometry-help-please/96704 "2001-12-03T22:06:14Z")
**Posts on this page:** 20
**Page:** 1
### Author:  [@Spavined\_Gelding](https://boards.straightdope.com/u/Spavined_Gelding)
#### Post date: [December 3, 2001, 10:06pm UTC](https://boards.straightdope.com/t/a-little-geometry-help-please/96704/1 "2001-12-03T22:06:14Z")
How do I divide a circle into 14 equal segments, that is 14 arcs of equal length and 14 equal radii angles? I’m trying to figure out how to design an old time wooden wagon wheel with 14 spokes.
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### Author:  [@lissener](https://boards.straightdope.com/u/lissener)
#### Post date: [December 3, 2001, 10:09pm UTC](https://boards.straightdope.com/t/a-little-geometry-help-please/96704/2 "2001-12-03T22:09:49Z")
On paper? Or do you have any drawing software? In Illustrator, for example, you can make a 14-spoke wheel by entering the right numbers in the right boxes.
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### Author:  [@friedo](https://boards.straightdope.com/u/friedo)
#### Post date: [December 3, 2001, 10:16pm UTC](https://boards.straightdope.com/t/a-little-geometry-help-please/96704/3 "2001-12-03T22:16:07Z")
Well, 360 / 14 is roughly 25.7. So draw a circle, and draw a radius pointing directly up. Then measure 25.7 degrees from that radius and draw another one. Repeat.
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### Author:  [@Crafter\_Man](https://boards.straightdope.com/u/Crafter_Man)
#### Post date: [December 3, 2001, 10:47pm UTC](https://boards.straightdope.com/t/a-little-geometry-help-please/96704/4 "2001-12-03T22:47:32Z")
Segment length = pi\*D/n, where
pi = 3.141592654
D = diameter of circle
n = number of segments
Example: If the wheel is 25 inches in diameter, and you have 14 segments, each segment would be pi\*25/14 = 5.610 inches long. Of course, this is measured along the circumference of the segment.
Is this what you’re asking for?
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### Author:  [@CalMeacham](https://boards.straightdope.com/u/CalMeacham)
#### Post date: [December 4, 2001, 3:10am UTC](https://boards.straightdope.com/t/a-little-geometry-help-please/96704/5 "2001-12-04T03:10:28Z")
The protractor is your friend.
As friedo said, 25.714 degrees per segment.
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### Author:  [@Arnold\_Winkelried](https://boards.straightdope.com/u/Arnold_Winkelried)
#### Post date: [December 4, 2001, 3:23am UTC](https://boards.straightdope.com/t/a-little-geometry-help-please/96704/6 "2001-12-04T03:23:13Z")
Shouldn’t that be 25.7142857142857… ? I would hate for his wheel to turn out elliptic.
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### Author:  [@RM\_Mentock](https://boards.straightdope.com/u/RM_Mentock)
#### Post date: [December 4, 2001, 9:25am UTC](https://boards.straightdope.com/t/a-little-geometry-help-please/96704/7 "2001-12-04T09:25:28Z")
> [@](#):
>
> \*Originally posted by Arnold Winkelried \*
> \*\*Shouldn’t that be 25.7142857142857… ? I would hate for his wheel to turn out elliptic. \*\*
Better make it 25 5/7 then.
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### Author:  [@23skidoo](https://boards.straightdope.com/u/23skidoo)
#### Post date: [December 4, 2001, 12:25pm UTC](https://boards.straightdope.com/t/a-little-geometry-help-please/96704/8 "2001-12-04T12:25:21Z")
> [@](#):
>
> \*Originally posted by lissener \*
> \*\*On paper? Or do you have any drawing software? In Illustrator, for example, you can make a 14-spoke wheel by entering the right numbers in the right boxes. \*\*
Microsoft Excel will work to. Put 14 equal values into the first column, then make a circle graph out of the data. If you make each section white, you should have what you need.
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### Author:  [@Big\_Sam](https://boards.straightdope.com/u/Big_Sam)
#### Post date: [December 4, 2001, 3:42pm UTC](https://boards.straightdope.com/t/a-little-geometry-help-please/96704/9 "2001-12-04T15:42:43Z")
> [@](#):
>
> \*Originally posted by Arnold Winkelried \*
> \*\*Shouldn’t that be 25.7142857142857… ? I would hate for his wheel to turn out elliptic. \*\*
Good luck holding that tolerance.
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### Author:  [@Ethilrist](https://boards.straightdope.com/u/Ethilrist)
#### Post date: [December 4, 2001, 3:45pm UTC](https://boards.straightdope.com/t/a-little-geometry-help-please/96704/10 "2001-12-04T15:45:10Z")
Make one with 16 and take out two.
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### Author:  [@Tenebras](https://boards.straightdope.com/u/Tenebras)
#### Post date: [December 4, 2001, 5:12pm UTC](https://boards.straightdope.com/t/a-little-geometry-help-please/96704/11 "2001-12-04T17:12:31Z")
Well, if you want to do it right, you’re going to need more than a straightedge and compass. Damn that heptagon inconstructability, damn it all to… er, sorry.
I recommend building your wheel with 17 spokes, spaced evenly. And then you can be all cool and [Gauss-like](http://www.math.niu.edu/~rusin/known-math/98/heptadecagon). Wouldn’t that make you feel warm and twitchy inside?
Tenebras
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### Author:  [@jmonster](https://boards.straightdope.com/u/jmonster)
#### Post date: [December 4, 2001, 5:53pm UTC](https://boards.straightdope.com/t/a-little-geometry-help-please/96704/12 "2001-12-04T17:53:53Z")
> [@](#):
>
> \*Originally posted by Ethilrist \*
> Make one with 16 and take out two.
It would be more symmetrical to make one with 28 and take out every other one.
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### Author:  [@erislover](https://boards.straightdope.com/u/erislover)
#### Post date: [December 4, 2001, 7:43pm UTC](https://boards.straightdope.com/t/a-little-geometry-help-please/96704/13 "2001-12-04T19:43:45Z")
Apart from simply using a protractor with a little algebra, there is no compass and straightedge solution. Since one can only bisect angles we would need to show that some multiple of seven is also a power of two. Thus:
7x = 2[sup]y[/sup]
ln(7x) = yln2
y = [ln(7x)]/[ln2]
No integral solutions.
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### Author:  [@ultrafilter](https://boards.straightdope.com/u/ultrafilter)
#### Post date: [December 4, 2001, 7:54pm UTC](https://boards.straightdope.com/t/a-little-geometry-help-please/96704/14 "2001-12-04T19:54:57Z")
> [@](#):
>
> \*Originally posted by erislover \*
> \*\*Apart from simply using a protractor with a little algebra, there is no compass and straightedge solution. Since one can only bisect angles we would need to show that some multiple of seven is also a power of two. Thus:
>
> 7x = 2[sup]y[/sup]
> ln(7x) = yln2
> y = [ln(7x)]/[ln2]
>
> No integral solutions. \*\*
The correction is true, but the method isn’t quite right. Otherwise, how do you explain my ability to construct a regular triangle?
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### Author:  [@ultrafilter](https://boards.straightdope.com/u/ultrafilter)
#### Post date: [December 4, 2001, 7:57pm UTC](https://boards.straightdope.com/t/a-little-geometry-help-please/96704/15 "2001-12-04T19:57:57Z")
Of course, it’s also necessary to actually show that there are no integer values of x that lead to ln(7x)/ln(2) being an integer. This is not obvious.
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### Author:  [@N9IWP](https://boards.straightdope.com/u/N9IWP)
#### Post date: [December 4, 2001, 9:26pm UTC](https://boards.straightdope.com/t/a-little-geometry-help-please/96704/16 "2001-12-04T21:26:06Z")
Arent you in general structurely better off with an odd number of spokes?
That way you dont have any spoke directly opposite another one. Though I coundn’t tell you why having opposite spokes is bad.
Brian
thinks he may have read that somewhere
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### Author:  [@Spavined\_Gelding](https://boards.straightdope.com/u/Spavined_Gelding)
#### Post date: [December 4, 2001, 9:56pm UTC](https://boards.straightdope.com/t/a-little-geometry-help-please/96704/17 "2001-12-04T21:56:19Z")
As a little fuller explanation, I am a dilettante weekend painter. I am trying to draw/paint a 14 spoke wagon wheel. One of my hang-ups is incorrect obvious detail. Wooden wagon wheels, at least big ones such as would be found on Civil War period artillery, is wooden hubbed with 14 spokes. I am looking for a straight edge and compass method for the spoke placement instead of the frustrating trial and adjust method I have been using.
I can’t believe that old time wheelwrights just measured off the circumference and knocked that into 14 segments. After all, the wheelwright had to build the hub too, and that is not the sort of thing that can be easily segmented. It may well be that they just used a jig that set a more or less 25 or 26 degree angle on the hub and extended that to the rim. In any event, it seems to be the consensus here that there is no simple straight edge and compass method for doing this.
Regarding the working geometry of a spoked wheel, on the old wooden wheel the weight on the axel was shared between the spokes running up and the spokes going down. The lower spoke supported some of the weight on the axel and part of the weight hung from the upper spoke. Thus, the spokes needed to be paired so as not to over strain any one spoke. In the wire spoked bicycle wheel, I think, the weight on the axel hangs from several adjacent spokes and little weight is held up by the lower spokes.
None-the-less, thanks for the help.
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### Author:  [@Cabbage](https://boards.straightdope.com/u/Cabbage)
#### Post date: [December 4, 2001, 10:50pm UTC](https://boards.straightdope.com/t/a-little-geometry-help-please/96704/18 "2001-12-04T22:50:27Z")
You can construct a regular heptagon with a compass and _marked_ straightedge, but I haven’t found how to do it on the internet. You may want to hunt up the article: _Angle Trisection, the Heptagon, and the Triskaidekagon_, by Andrew M. Gleason, in _The American Mathematical Monthly_ vol. 95 #3 (March 1988), pp. 185-194.
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### Author:  [@keeper0](https://boards.straightdope.com/u/keeper0)
#### Post date: [December 4, 2001, 11:59pm UTC](https://boards.straightdope.com/t/a-little-geometry-help-please/96704/19 "2001-12-04T23:59:25Z")
Since you are drawing it, a compass and a ruler will probably be easier than a protractor.
One of the earlier posts was close, but I think it should be made clearer.
Draw a circle of desired radius with a compass. For the moment assume a radius of 10.
Now set your compass to 0.445 of your original radius. In our example: 4.45.
Mark one point on your edge. Put the pivot of the compass on that point and make a new mark on the edge with the pencil side of the compass. Put the pivot on the new mark and make a third mark with the pencil.
This should step you around the circle in 14 increments sufficiently precise. You will surely be able to compensate for small errors with artistic talent.
(Note the .445 is NOT 1/14th of the circumference, it is the length of the side of the heptagon circumscribed by the circle.)
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### Author:  [@RM\_Mentock](https://boards.straightdope.com/u/RM_Mentock)
#### Post date: [December 5, 2001, 1:21am UTC](https://boards.straightdope.com/t/a-little-geometry-help-please/96704/20 "2001-12-05T01:21:05Z")
> [@](#):
>
> \*Originally posted by Cabbage \*
> \*\*You can construct a regular heptagon with a compass and _marked_ straightedge, but I haven’t found how to do it on the internet. You may want to hunt up the article: _Angle Trisection, the Heptagon, and the Triskaidekagon_, by Andrew M. Gleason, in _The American Mathematical Monthly_ vol. 95 #3 (March 1988), pp. 185-194. \*\*
[Here’s a nice angle trisection with compass and straightedge, no mark.](http://mentock.home.mindspring.com/trisect.htm)🙂
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